The same open sets

Question

How to prove that the two distances $d(x,y)=|x-y|$ and $d'(x,y)= |\exp(x)-\exp(y)|$ induced the same topology on R ?

I know that for $a\in R, r>0$

$B_d(a,r)=]a-r, a+r[$,

$B_{d'}(a,r)=]\ln(\exp(a)-r), \ln(\exp(a)+r)[$

How to do ?

Answer 1

I think that you can use this fact:

If $X$ is a first countable space then a function $f: X\to Y$ is continuos if and only if for each convergent sequence $\{x_n\}_n$ convergent to a point $x$ then $\{f(x_n)\}_n$ is convergent to $f(x)$

Now you can observe that if $X$ has two Topology $\tau_1$ and $\tau_2$ such that $(X,\tau_1)$ is first countable and for each $\tau_1-$convergent sequence $\{x_n\}_n$ to a point $x$ is $\tau_2-$ convergent sequence to the point $x$ then $\tau_2\subseteq \tau_1$ infact by first lemma you have that the identity map

$Id: (X,\tau_1)\to (X,\tau_2)$

is a countinuos function so for each open set $A\in \tau_2$ then $Id^{-1}(A)=A$ is open in $\tau_1$. So $\tau_2\subseteq \tau_1$.

In your case $\mathbb{R}$ is a first countable set with respect both Topology because they are induced by a distance. You have also that the function $exp$ is a countinuos function from $\mathbb{R}$ to $\mathbb{R}$ so if $\{x_n\}_n$ is convergent to $x$ with respect $d_1$ then

$\lim_{n\to \infty}|exp(x_n)-exp(x)|=|exp(x)-exp(x)|=0$

so $\{x_n\}$ it is convergent to $x$ with respect to $d_2$ so you have that

$\tau_2\subseteq \tau_1$

Now you can hypothesize that a sequence $\{x_n\}_n$ is convergent with respect $d_2$ to a point $x$ then

$\{x_n\}_n$ is convergent with respect $d_1$ to a point $x$ because $log$ is a countinuos function and so

$\lim_{n\to \infty}|x_n-x|=\lim_{n\to\infty}| log(e^{x_n})-log(e^x)|=0$

So you have that $\tau_1\subseteq \tau_2$

It is obvious that you can generalize your statement:

If $f:\mathbb{R}\to \mathbb{R}$ is an injective function then it is possible define a distance $d_2$ on $\mathbb{R}$ in the following way:

$d_2(x,y):=|f(x)-f(y)|$

If $f:(\mathbb{R},\tau_{eu})\to (\mathbb{R},\tau_{eu})$ is a continuos function then $\tau_2\subseteq \tau_{eu}$.

If $f$ in bijective and your inverse it is continuos then

$\tau_{eu}\subseteq \tau_2$

Answer 2

Suppose that $(a, b)$ is an open ball in the ordinary topology. (That is, it is the ball $$ B_d\left(\frac{a+b}2, \frac{b-a}2\right), $$ but that doesn't matter for our argument.) Then take $$ x = \log\left(\frac{\exp(a)+\exp(b)}2\right), r = \frac{\exp(b) - \exp(a)}{2}. $$ We have that $$ B_{d'}(x, r) = \left(\log(\exp(x) - r), \log(\exp(x) + r)\right) = (a, b), $$ where I've used your expression for $B_{d'}(x, r)$ in your question. Conversely, consider $B_{d'}(x, r)$ for arbitrary $x, r > 0$. We consider two cases. If $\exp(x) - r$ is positive, then we have your expression again, namely $$ B_{d'}(x, r) = \left(\log(\exp(x) - r), \log(\exp(x) + r)\right), $$ and then we already have an open interval in the normal sense. On the other hand, if $\exp(x) - r \leq 0$, then we have that $$ B_{d'}(x, r) = (-\infty, \log(\exp(x) + r)). $$ Can you prove that this set is open in the ordinary topology (induced by $d$)?